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Homework _6 Solutions Final

Homework _6 Solutions Final - ChemE 178 Homework#6...

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ChemE 178: Homework #6 Solutions 1.) Polymer blends are to be made from a poly(ethylenepropylene) (h-PEP) of weight average molecular weight Mw = 175,000, Mw/Mn=1.01 and a perdeuterated- poly(ethylenepropylene) (d-PEP), Mw=200,000, Mw/Mn=1.01. The d-PEP is the same in all respect as the h-PEP except that the d-PEP has all hydrogens exchanged for deuterium. The Flory parameter χ for the d-PEP/h-PEP pair is given by χ = (0.571/T)-6.6x10 -4 , where T is the absolute temperature (K). Compute and plot the following: _______________________________________________________________________ _ A.) The normalized free energy of mixing G mix /Mk B T versus ϕ d-PB at the following temperatures: 300K, 350K, 385K and 450K. Use a spreadsheet program like Excel or a plotting program with computational capability to produce these plots. Use scales so that important details are visible. Begin with the Flory-Huggins Equation: ( 29 ( 29 ln ln A B A B A B B A B G Mk T N N φ φ φ φ χφ φ = + + To get N A and N B , you will need to use the number average molecular weight. d-PEP: 200,000 198,020 1.01 w n M M PDI = = = h-PEP: 175,000 173,267 1.01 w n M M PDI = = = Assuming that the h-PEP and d-PEP have similar monomeric volumes, the segment lengths are simply the degree of polymerization. Note: This is not always the case. 173,267 2,475 70 198,020 2,475 80 h PEP A d PEP B N N N N - - = = = = = = The polymer blend is incompressible, so 1 A B φ φ = + Given the expression for χ, you can use the Flory-Huggins equation to plot the normalized free energy as a function of volume fraction.
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Gibbs Energy of Mixing -1.40E-04 -1.20E-04 -1.00E-04 -8.00E-05 -6.00E-05 -4.00E-05 -2.00E-05 0.00E+00 2.00E-05 4.00E-05 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 φ (d-PB) G/MkT 300 K 350 K 385 K 450 K B.) Compute the binodal and spinodal ϕ 's as a function of temperature and use these to compute and plot the phase diagram for this system. The binodal is define as ( 29 ( 29 ( 29 1 2 1 1 ln 1 ln 1 1 2 A A mix mix A B A A A A B G G N N φ φ φ φ χ φ φ φ ∂∆ ∂∆ = = + - + + - ÷ ÷ The binodal is calculated using the condition that binodal point lies on the same tangent line. For this problem, the free energy curves are symmetric (N A = N B ), so the common tangent line is horizontal (a slope of zero). As a result, the binodal point is the local minimum of the free energy curve.
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Homework _6 Solutions Final - ChemE 178 Homework#6...

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